Did you know that half of New Yorkers are below the median income?

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Originally posted by GilesGuthrie
What about the other 1%? Dolt.

Also, it's not always required that someone is on the median income. The median is an arbitrary point in a dataset, and it is not required that there be any members of that dataset on the actual median point.

You are therefore a total loser.

Hahaha.

Make that 49.5 percent above and below.

There must be at least a few people on the median point, no?
 
Originally posted by Sage
As long as we're talking about this stuff, I have a question...

When you find standard deviation, how come you square (then un-square) the deviation values instead of just taking the absolute value?

Because squaring makes all the values unsigned, which is vital for SD calculations.
 
But taking the absolute value has the same effect, doesn't it?
 
Originally posted by Klostrophobic
Hahaha.

Make that 49.5 percent above and below.

There must be at least a few people on the median point, no?

Not necessarily, but in a population set the size of New York City, it's a pretty safe bet.
 
Originally posted by GilesGuthrie
No, the modal value in a dataset is the value that has the highest individual population.

So, in the dataset 1, 3, 3, 5, 5, 5, 5, 5, 6, 7, 8, 9, 9, 9, 24523, the modal value is 5, but the average is 1640.2.
The mode and the Median are both averages, average isnt a mathematical term. And if those numbers were salaries half of the population wouldnt be below the average if average was interpreted to mean the income bracket with the largest number of people. It works better with a large set and you get a bell curve.
 
all I'm really hunting for i an outfit that looks good, sn00pie, great taste in music. I thought I had to mention it, and yer. I had a maths exam today, and I was using my sister calulator and was doing standard deviation, and I dont know how to do it on hers so I had to do it using the formula which I had (luckily) memorised.
 

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